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Let $\mu$ be a Borel measure in $\mathbb{R}$ such that $\mu(I) \leq v^a(I)$ for each bounded interval $I$, where $a>1$. Show that $\mu=0$.

($v(R)$ is the volume of $R$)

Do we maybe use the following to show that?

For each rectangle $R$, $m^*(R)=v(R)$.

Or do I have to do it otherwise?? Could you give me some hints??

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    $\begingroup$ Could you say what is $v$? $\endgroup$ – Petite Etincelle Oct 14 '14 at 23:09
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    $\begingroup$ It should be $\mu(I) \le (v(I))^a$, where $v(\cdot)$ is the Lebesgue measure. $\endgroup$ – Orest Bucicovschi Oct 14 '14 at 23:13
  • $\begingroup$ @LiuGang $v(R)$ is the volume of $R$ $\endgroup$ – Mary Star Oct 14 '14 at 23:22
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For every interval $I =[x,y)$ and every $n \in \mathbb{N}$ we have that $$I= [x,y)= \bigcup_{k = 0}^{n-1} \underbrace{\left[ x + (y-x)\frac{k}{n},x + (y-x)\frac{k+1}{n}\right)}_{I_k}$$ Then $$\mu(I_k) \leq \nu(I_k)^a = \left(\frac{y-x}{n}\right)^a$$ Now by additivity we have $$\mu(I) = \sum_{k=0}^{n-1} \mu(I_k) \leq \sum_{k=0}^{n-1}\left(\frac{y-x}{n}\right)^a = (y-x)^a n^{1-a} \stackrel{n \to \infty}{\longrightarrow} 0 $$ Hence $\mu(I)=0$ for every interval so $\mu = 0$.

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  • $\begingroup$ I understand!! Thank you very much!!! :-) $\endgroup$ – Mary Star Oct 24 '14 at 15:42
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Hint: Notice that $[a,a+b)=[a,a+b/2)\cup [a+b/2,a+b)$. Use this to show that $\mu([a,a+b))$ is smaller than any $\varepsilon>0$.

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  • $\begingroup$ Ok!!! Thanks a lot!!! :-) $\endgroup$ – Mary Star Oct 24 '14 at 15:44

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